MN - средняя линия треугольника АРС
(АМ=МС)
Значит,
PN=NC
BP:PN=7:3
BP:BN=7:10
BP:BC=7:13
Пусть
S_( Δ ABC)=2s
тогда
S_( ΔMBC)=s ( основание MC=(1/2)AC, высота общая)
S_( ΔAPC):S_( ΔABC)=PC:BC=6:7
S_( ΔAPC)=6s/7
S_(MNC)=(1/4)S_( ΔAPC)=6s/28=3s/14
S_( ΔBMN)=S_( ΔMBC)-S_(MNC)=s-(3s/14)=11s/14
S_(ΔBKP):S_( ΔBMN)=7^2:10^2
[b]S_(ΔBKP)[/b]= 77s/200
S_(KPCM)=S_( Δ MBC)- S_( Δ BKP)=s-(77s/200)=123s/200
S_(ΔBKP):S_(KPCM)=77:123